Exercise 1, § 4

Tag 011G, permanent. Cite it as [Bourbaki, Tag 011G].

Lie Groups and Lie Algebras, VIII, 231

Retain the notations of no. 2. Let $\lambda \in k^B$ . Associate to any $\alpha \in B$ the endomorphisms $X_{-\alpha}^{\lambda}, H_{\alpha}^{\lambda}, X_{\alpha}^{\lambda}$ of the vector space E such that

X_{-\alpha}^{\lambda}(\alpha_1, . . . , \alpha_n) = (\alpha , \alpha_1, . . . , \alpha_n)

H_{\alpha}^{\lambda}(\alpha_1, . . . , \alpha_n) = (\lambda (\alpha )-\sum_{i=1}^nn(\alpha_i, \alpha ))(\alpha_1, . . . , \alpha_n)

The vector $X_{\alpha}^{\lambda}(\alpha_1, . . . , \alpha_n)$ is defined by induction on $n$ by the formula

X_{\alpha}^{\lambda}(\alpha_1, . . . , \alpha_n) = (X_{-\alpha_1}^{\alpha}X_{\alpha}^{\lambda}-\delta_{\alpha ,\alpha_1}H_{\alpha}^{\lambda})(\alpha_2, . . . , \alpha_n)

where $\delta_{\alpha ,\alpha_1}$ is the Kronecker symbol; we agree that, if $(\alpha_1, . . . , \alpha_n)$ is the empty word, then $X_{\alpha}^{\lambda}(\alpha_1, . . . , \alpha_n)$ is zero.

Show that Lemmas 1 and 2 remain true for these endomorphisms. We thus obtain a representation $\rho_{\lambda}:\mathfrak{a}\rightarrow \mathfrak{g}\mathfrak{l}(E)$ such that

\rho_{\lambda}(x_{\alpha}) =X_{\alpha}^{\lambda}, \rho_{\lambda}(h_{\alpha}) =H_{\alpha}^{\lambda}, \rho_{\lambda}(x_{-\alpha}) =X_{-\alpha}^{\lambda}

Solution

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Cited by

Exercise 1, § 6

Cites

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